Question: Consider the parametric curve: $\begin{aligned} x&=\dfrac{-5}{t^2} \\\\ y&=e^{-t} \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=-5$ to $t=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{-5}^{0} \sqrt{\dfrac{100}{t^6}+e^{-2t}}\,dt$ (Choice B) B $\int_{-5}^{0} \sqrt{\dfrac{100}{t^9}+e^{t^2}}\,dt$ (Choice C) C $\int_{-5}^{0} \sqrt{\dfrac{100}{t^9}-e^{t^2}}\,dt$ (Choice D) D $\int_{-5}^{0} \sqrt{\dfrac{100}{t^6}-e^{-2t}}\,dt$
This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\dfrac{-5}{t^2}\right] \\\\ &=\dfrac{10}{t^3} \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[e^{-t}\right] \\\\ &=-e^{-t} \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{-5}^{0} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{-5}^{0} \sqrt{\left(\dfrac{10}{t^3}\right)^2+\left(-e^{-t}\right)^2}\,dt \\\\ &=\int_{-5}^{0} \sqrt{\dfrac{100}{t^6}+e^{-2t}}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=-5$ to $t=0$ : $\int_{-5}^{0} \sqrt{\dfrac{100}{t^6}+e^{-2t}}\,dt$